How to show unbiased estimator of combination of bernoulli and normal variables?

Question

$X_1, X_2, \ldots, X_n$ is a random sample from $\mathrm{Bernoulli}(\theta)$, $\epsilon_1, \epsilon_2, \ldots, \epsilon_n$ are independent $\mathcal N(0, \sigma^2)$, independent of $X_i$.

Define $Y_i = \theta X_i + \epsilon_i$, for $i = 1, 2, \ldots,n$. Define estimating function $$ \psi[\theta;(X,Y)] = \sum_{i =1}^n(Y_i - \theta X_i) \>. $$

a) Show that $\psi[\theta;(X,Y)]$ is unbiased estimator of $\theta$ if $\mathbb E(\psi[\theta;(X,Y)])=0$.

b) Find the estimator $\hat\theta$ such that $\psi[\hat\theta;(X,Y)] = 0$. Is $\hat\theta$ unbiased?

Answer 1

I'm still not quite sure how part (a) is different from part (b) but, from your comment above it appears you are now only asking about part (b), so:

If $\psi[\hat{\theta};(X,Y)] = 0$, then

$$ \sum_{i =1}^n(Y_i - \hat{\theta} X_i) \> = 0 $$

So,

$$ \sum_{i=1}^{n} Y_i = \hat{\theta} \cdot \sum_{i=1}^{n} X_i $$

Therefore $\hat{\theta} = \overline{Y}/\overline{X}$, the ratio of the sample means, satisfies $\psi[\hat{\theta};(X,Y)] = 0$. Regarding unbiasedness, it is easy to see that

$$ E( \hat{\theta} ) = E\left( \frac{ \sum_{i=1}^{n} \theta X_i + \varepsilon_{i} }{ \sum_{i=1}^{n} X_i }\right) = \theta + E \left( \frac{ \sum_{i=1}^{n} \varepsilon_i }{ \sum_{i=1}^{n} X_i }\right), $$

Edit: Based on the discussion in the comments, I've edited my answer. Let $B=\sum_{i=1}^{n} X_{i}$ and $Z = \sum_{i=1}^{n} \varepsilon_i$. Then, $B \sim {\rm Binomial}(n,\theta)$ and $Z \sim N(0, n \sigma^{2})$. Since the errors are independent of the $X_{i}$,

$$ E \left( \frac{ \sum_{i=1}^{n} \varepsilon_i }{ \sum_{i=1}^{n} X_i }\right) = E(Z) \cdot E \left( \frac{1}{B} \right) $$

Clearly $E(Z) = 0$. Assuming $\theta < 1$, $P(B = 0) = (1-\theta)^{n} > 0$. Therefore $E \left( \frac{1}{B} \right) = \infty$, so $E \left( \frac{ \sum_{i=1}^{n} \varepsilon_i }{ \sum_{i=1}^{n} X_i }\right)$ doesn't exist. Therefore $E(\hat{\theta})$ doesn't exist whenever $\theta < 1$, so $\hat{\theta}$ can't be unbiased (although it is consistent as long as $\theta > 0$).